
Find integration of :- \[\int {\cos x\log \left( {\tan \dfrac{x}{2}} \right)} dx = \]
(a) \[\sin x\log \left| {\tan x} \right| - x + C\]
(b) \[ - \sin x\log \left| {\tan \dfrac{x}{2}} \right| + x + C\]
(c) \[ - \sin x\log \left| {\tan \dfrac{x}{2}} \right| - x + C\]
(d) \[\sin x\log \left| {\tan \dfrac{x}{2}} \right| - x + C\]
Answer
553.5k+ views
Hint: Here, we need to find the value of the given integral. We will use integration by parts and trigonometric identities to simplify the given equation and find the integral of the given function. Integration is a method by which we can find summation of discrete data.
Formula Used: We will use the following formula:
1.Using integration by parts, the integral of the product of two differentiable functions of \[x\] can be written as \[\int {uv} dx = u\int v dx - \int {\left( {\dfrac{{d\left( u \right)}}{{dx}} \times \int v dx} \right)dx} \], where \[u\] and \[v\] are the differentiable functions of \[x\].
2.The tangent of an angle \[\theta \] can be written as \[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\].
3.The secant of an angle \[\theta \] can be written as \[\sec \theta = \dfrac{1}{{\cos \theta }}\].
Complete step-by-step answer:
We will integrate the simplified function using integration by parts.
Rewriting the equation, we get
\[ \Rightarrow \int {\cos x\log \left( {\tan \dfrac{x}{2}} \right)} dx = \int {\cos x \times \log \left( {\tan \dfrac{x}{2}} \right)} dx\]
Let \[u\] be \[\log \left( {\tan \dfrac{x}{2}} \right)\] and \[v\] be \[\cos x\].
Therefore, by integrating \[\int {\cos x \times \log \left( {\tan \dfrac{x}{2}} \right)} dx\] by parts, we get
\[ \Rightarrow \int {\cos x \times \log \left( {\tan \dfrac{x}{2}} \right)} dx = \log \left( {\tan \dfrac{x}{2}} \right)\int {\left( {\cos x} \right)} dx - \int {\left[ {\dfrac{{d\left\{ {\log \left( {\tan \dfrac{x}{2}} \right)} \right\}}}{{dx}} \times \int {\left( {\cos x} \right)} dx} \right]} dx\]
The derivative of a function of the form \[\log \left\{ {f\left( x \right)} \right\}\] is \[\dfrac{1}{{f\left( x \right)}} \times \dfrac{{d\left\{ {f\left( x \right)} \right\}}}{{dx}}\].
Thus, we get
\[ \Rightarrow \int {\cos x \times \log \left( {\tan \dfrac{x}{2}} \right)} dx = \log \left( {\tan \dfrac{x}{2}} \right)\int {\left( {\cos x} \right)} dx - \int {\left[ {\dfrac{1}{{\tan \dfrac{x}{2}}} \times \dfrac{{d\left( {\tan \dfrac{x}{2}} \right)}}{{dx}} \times \int {\left( {\cos x} \right)} dx} \right]} dx\]
Simplifying the derivative, we get
\[ \Rightarrow \int {\cos x \times \log \left( {\tan \dfrac{x}{2}} \right)} dx = \log \left( {\tan \dfrac{x}{2}} \right)\int {\left( {\cos x} \right)} dx - \int {\left[ {\dfrac{1}{{\tan \dfrac{x}{2}}} \times {{\sec }^2}\dfrac{x}{2} \times \dfrac{{d\left( {\dfrac{x}{2}} \right)}}{{dx}} \times \int {\left( {\cos x} \right)} dx} \right]} dx\]
Therefore, we get
\[ \Rightarrow \int {\cos x \times \log \left( {\tan \dfrac{x}{2}} \right)} dx = \log \left( {\tan \dfrac{x}{2}} \right)\int {\left( {\cos x} \right)} dx - \int {\left[ {\dfrac{1}{{\tan \dfrac{x}{2}}} \times {{\sec }^2}\dfrac{x}{2} \times \dfrac{1}{2} \times \int {\left( {\cos x} \right)} dx} \right]} dx\]
The tangent of an angle \[\theta \] can be written as \[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\].
The secant of an angle \[\theta \] can be written as \[\sec \theta = \dfrac{1}{{\cos \theta }}\].
Rewriting the terms in the parentheses in terms of sine and cosine, we get
\[ \Rightarrow \int {\cos x \times \log \left( {\tan \dfrac{x}{2}} \right)} dx = \log \left( {\tan \dfrac{x}{2}} \right)\int {\left( {\cos x} \right)} dx - \int {\left[ {\dfrac{1}{{\dfrac{{\sin \dfrac{x}{2}}}{{\cos \dfrac{x}{2}}}}} \times \dfrac{1}{{{{\cos }^2}\dfrac{x}{2}}} \times \dfrac{1}{2} \times \int {\left( {\cos x} \right)} dx} \right]} dx\]
Simplifying the expression, we get
\[\begin{array}{l} \Rightarrow \int {\cos x \times \log \left( {\tan \dfrac{x}{2}} \right)} dx = \log \left( {\tan \dfrac{x}{2}} \right)\int {\left( {\cos x} \right)} dx - \int {\left[ {\dfrac{{\cos \dfrac{x}{2}}}{{\sin \dfrac{x}{2}}} \times \dfrac{1}{{{{\cos }^2}\dfrac{x}{2}}} \times \dfrac{1}{2} \times \int {\left( {\cos x} \right)} dx} \right]} dx\\ \Rightarrow \int {\cos x \times \log \left( {\tan \dfrac{x}{2}} \right)} dx = \log \left( {\tan \dfrac{x}{2}} \right)\int {\left( {\cos x} \right)} dx - \int {\left[ {\dfrac{1}{2} \times \dfrac{1}{{\sin \dfrac{x}{2}}} \times \dfrac{1}{{\cos \dfrac{x}{2}}} \times \int {\left( {\cos x} \right)} dx} \right]} dx\end{array}\]
Therefore, we get
\[ \Rightarrow \int {\cos x \times \log \left( {\tan \dfrac{x}{2}} \right)} dx = \log \left( {\tan \dfrac{x}{2}} \right)\int {\left( {\cos x} \right)} dx - \int {\left[ {\dfrac{1}{{2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}} \times \int {\left( {\cos x} \right)} dx} \right]} dx\]
Applying the trigonometric identity \[2\sin A\cos A = \sin 2A\], we get
\[ \Rightarrow \int {\cos x \times \log \left( {\tan \dfrac{x}{2}} \right)} dx = \log \left( {\tan \dfrac{x}{2}} \right)\int {\left( {\cos x} \right)} dx - \int {\left[ {\dfrac{1}{{\sin x}} \times \int {\left( {\cos x} \right)} dx} \right]} dx\]
Now, the integral of \[\cos x\] is \[\sin x\].
Substituting \[\int {\cos x} dx = \sin x\] in the equation, we get
\[ \Rightarrow \int {\cos x \times \log \left( {\tan \dfrac{x}{2}} \right)} dx = \log \left( {\tan \dfrac{x}{2}} \right)\sin x - \int {\left[ {\dfrac{1}{{\sin x}} \times \sin x} \right]} dx\]
Simplifying the expression, we get
\[ \Rightarrow \int {\cos x \times \log \left( {\tan \dfrac{x}{2}} \right)} dx = \sin x\log \left( {\tan \dfrac{x}{2}} \right) - \int {\left( 1 \right)} dx\]
The integral of a constant with respect to a variable is the product of the constant and the variable.
Therefore, we get
\[ \Rightarrow \int {\cos x\log \left( {\tan \dfrac{x}{2}} \right)} dx = \sin x\log \left( {\tan \dfrac{x}{2}} \right) - x + C\], where \[C\] is the constant of integration
Therefore, we get the value of the integral \[\int {\cos x\log \left( {\tan \dfrac{x}{2}} \right)} dx\] as \[\sin x\log \left( {\tan \dfrac{x}{2}} \right) - x + C\], where \[C\] is the constant of integration.
Thus, the correct option is option (d).
Note: We simplified \[\dfrac{{d\left( {\tan \dfrac{x}{2}} \right)}}{{dx}}\] as \[{\sec ^2}\dfrac{x}{2} \times \dfrac{{d\left( {\dfrac{x}{2}} \right)}}{{dx}}\]. This is because the derivative of a function of the form \[\tan \left\{ {f\left( x \right)} \right\}\] is \[{\sec ^2}\left\{ {f\left( x \right)} \right\} \times \dfrac{{d\left\{ {f\left( x \right)} \right\}}}{{dx}}\].
We simplified \[\dfrac{{d\left( {\dfrac{x}{2}} \right)}}{{dx}}\] as \[\dfrac{1}{2}\]. This is because the derivative of a function of the form \[af\left( x \right)\] is \[a\dfrac{{d\left\{ {f\left( x \right)} \right\}}}{{dx}}\], and the derivative of a variable with respect to itself is 1.
We used the trigonometric identity \[2\sin A\cos A = \sin 2A\]. This is the trigonometric identity for the sine of the double of an angle.
Formula Used: We will use the following formula:
1.Using integration by parts, the integral of the product of two differentiable functions of \[x\] can be written as \[\int {uv} dx = u\int v dx - \int {\left( {\dfrac{{d\left( u \right)}}{{dx}} \times \int v dx} \right)dx} \], where \[u\] and \[v\] are the differentiable functions of \[x\].
2.The tangent of an angle \[\theta \] can be written as \[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\].
3.The secant of an angle \[\theta \] can be written as \[\sec \theta = \dfrac{1}{{\cos \theta }}\].
Complete step-by-step answer:
We will integrate the simplified function using integration by parts.
Rewriting the equation, we get
\[ \Rightarrow \int {\cos x\log \left( {\tan \dfrac{x}{2}} \right)} dx = \int {\cos x \times \log \left( {\tan \dfrac{x}{2}} \right)} dx\]
Let \[u\] be \[\log \left( {\tan \dfrac{x}{2}} \right)\] and \[v\] be \[\cos x\].
Therefore, by integrating \[\int {\cos x \times \log \left( {\tan \dfrac{x}{2}} \right)} dx\] by parts, we get
\[ \Rightarrow \int {\cos x \times \log \left( {\tan \dfrac{x}{2}} \right)} dx = \log \left( {\tan \dfrac{x}{2}} \right)\int {\left( {\cos x} \right)} dx - \int {\left[ {\dfrac{{d\left\{ {\log \left( {\tan \dfrac{x}{2}} \right)} \right\}}}{{dx}} \times \int {\left( {\cos x} \right)} dx} \right]} dx\]
The derivative of a function of the form \[\log \left\{ {f\left( x \right)} \right\}\] is \[\dfrac{1}{{f\left( x \right)}} \times \dfrac{{d\left\{ {f\left( x \right)} \right\}}}{{dx}}\].
Thus, we get
\[ \Rightarrow \int {\cos x \times \log \left( {\tan \dfrac{x}{2}} \right)} dx = \log \left( {\tan \dfrac{x}{2}} \right)\int {\left( {\cos x} \right)} dx - \int {\left[ {\dfrac{1}{{\tan \dfrac{x}{2}}} \times \dfrac{{d\left( {\tan \dfrac{x}{2}} \right)}}{{dx}} \times \int {\left( {\cos x} \right)} dx} \right]} dx\]
Simplifying the derivative, we get
\[ \Rightarrow \int {\cos x \times \log \left( {\tan \dfrac{x}{2}} \right)} dx = \log \left( {\tan \dfrac{x}{2}} \right)\int {\left( {\cos x} \right)} dx - \int {\left[ {\dfrac{1}{{\tan \dfrac{x}{2}}} \times {{\sec }^2}\dfrac{x}{2} \times \dfrac{{d\left( {\dfrac{x}{2}} \right)}}{{dx}} \times \int {\left( {\cos x} \right)} dx} \right]} dx\]
Therefore, we get
\[ \Rightarrow \int {\cos x \times \log \left( {\tan \dfrac{x}{2}} \right)} dx = \log \left( {\tan \dfrac{x}{2}} \right)\int {\left( {\cos x} \right)} dx - \int {\left[ {\dfrac{1}{{\tan \dfrac{x}{2}}} \times {{\sec }^2}\dfrac{x}{2} \times \dfrac{1}{2} \times \int {\left( {\cos x} \right)} dx} \right]} dx\]
The tangent of an angle \[\theta \] can be written as \[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\].
The secant of an angle \[\theta \] can be written as \[\sec \theta = \dfrac{1}{{\cos \theta }}\].
Rewriting the terms in the parentheses in terms of sine and cosine, we get
\[ \Rightarrow \int {\cos x \times \log \left( {\tan \dfrac{x}{2}} \right)} dx = \log \left( {\tan \dfrac{x}{2}} \right)\int {\left( {\cos x} \right)} dx - \int {\left[ {\dfrac{1}{{\dfrac{{\sin \dfrac{x}{2}}}{{\cos \dfrac{x}{2}}}}} \times \dfrac{1}{{{{\cos }^2}\dfrac{x}{2}}} \times \dfrac{1}{2} \times \int {\left( {\cos x} \right)} dx} \right]} dx\]
Simplifying the expression, we get
\[\begin{array}{l} \Rightarrow \int {\cos x \times \log \left( {\tan \dfrac{x}{2}} \right)} dx = \log \left( {\tan \dfrac{x}{2}} \right)\int {\left( {\cos x} \right)} dx - \int {\left[ {\dfrac{{\cos \dfrac{x}{2}}}{{\sin \dfrac{x}{2}}} \times \dfrac{1}{{{{\cos }^2}\dfrac{x}{2}}} \times \dfrac{1}{2} \times \int {\left( {\cos x} \right)} dx} \right]} dx\\ \Rightarrow \int {\cos x \times \log \left( {\tan \dfrac{x}{2}} \right)} dx = \log \left( {\tan \dfrac{x}{2}} \right)\int {\left( {\cos x} \right)} dx - \int {\left[ {\dfrac{1}{2} \times \dfrac{1}{{\sin \dfrac{x}{2}}} \times \dfrac{1}{{\cos \dfrac{x}{2}}} \times \int {\left( {\cos x} \right)} dx} \right]} dx\end{array}\]
Therefore, we get
\[ \Rightarrow \int {\cos x \times \log \left( {\tan \dfrac{x}{2}} \right)} dx = \log \left( {\tan \dfrac{x}{2}} \right)\int {\left( {\cos x} \right)} dx - \int {\left[ {\dfrac{1}{{2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}} \times \int {\left( {\cos x} \right)} dx} \right]} dx\]
Applying the trigonometric identity \[2\sin A\cos A = \sin 2A\], we get
\[ \Rightarrow \int {\cos x \times \log \left( {\tan \dfrac{x}{2}} \right)} dx = \log \left( {\tan \dfrac{x}{2}} \right)\int {\left( {\cos x} \right)} dx - \int {\left[ {\dfrac{1}{{\sin x}} \times \int {\left( {\cos x} \right)} dx} \right]} dx\]
Now, the integral of \[\cos x\] is \[\sin x\].
Substituting \[\int {\cos x} dx = \sin x\] in the equation, we get
\[ \Rightarrow \int {\cos x \times \log \left( {\tan \dfrac{x}{2}} \right)} dx = \log \left( {\tan \dfrac{x}{2}} \right)\sin x - \int {\left[ {\dfrac{1}{{\sin x}} \times \sin x} \right]} dx\]
Simplifying the expression, we get
\[ \Rightarrow \int {\cos x \times \log \left( {\tan \dfrac{x}{2}} \right)} dx = \sin x\log \left( {\tan \dfrac{x}{2}} \right) - \int {\left( 1 \right)} dx\]
The integral of a constant with respect to a variable is the product of the constant and the variable.
Therefore, we get
\[ \Rightarrow \int {\cos x\log \left( {\tan \dfrac{x}{2}} \right)} dx = \sin x\log \left( {\tan \dfrac{x}{2}} \right) - x + C\], where \[C\] is the constant of integration
Therefore, we get the value of the integral \[\int {\cos x\log \left( {\tan \dfrac{x}{2}} \right)} dx\] as \[\sin x\log \left( {\tan \dfrac{x}{2}} \right) - x + C\], where \[C\] is the constant of integration.
Thus, the correct option is option (d).
Note: We simplified \[\dfrac{{d\left( {\tan \dfrac{x}{2}} \right)}}{{dx}}\] as \[{\sec ^2}\dfrac{x}{2} \times \dfrac{{d\left( {\dfrac{x}{2}} \right)}}{{dx}}\]. This is because the derivative of a function of the form \[\tan \left\{ {f\left( x \right)} \right\}\] is \[{\sec ^2}\left\{ {f\left( x \right)} \right\} \times \dfrac{{d\left\{ {f\left( x \right)} \right\}}}{{dx}}\].
We simplified \[\dfrac{{d\left( {\dfrac{x}{2}} \right)}}{{dx}}\] as \[\dfrac{1}{2}\]. This is because the derivative of a function of the form \[af\left( x \right)\] is \[a\dfrac{{d\left\{ {f\left( x \right)} \right\}}}{{dx}}\], and the derivative of a variable with respect to itself is 1.
We used the trigonometric identity \[2\sin A\cos A = \sin 2A\]. This is the trigonometric identity for the sine of the double of an angle.
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Basicity of sulphurous acid and sulphuric acid are

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Trending doubts
What are the major means of transport Explain each class 12 social science CBSE

Which are the Top 10 Largest Countries of the World?

Draw a labelled sketch of the human eye class 12 physics CBSE

How much time does it take to bleed after eating p class 12 biology CBSE

Explain sex determination in humans with line diag class 12 biology CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

